Toán (1-1/x+1)(1-1/x+2)(1-1/x+3)…(1-1/x+2019) giúp mik nha 04/12/2021 By Hailey (1-1/x+1)(1-1/x+2)(1-1/x+3)…(1-1/x+2019) giúp mik nha
Đáp án: `(1 – 1/(x+1))(1-1/(x+2))(1-1/(x+3))…(1-1/(x+2019))` `= x/(x+1) . (x+1)/(x+2) . (x+2)/(x+3) … (x + 2018)/(x+2019)` `= [x . (x + 1)(x + 2)…(x+2018)]/[(x+1)(x+2)…(x+2018) . (x+2019)]` `= x/(x + 2019)` Giải thích các bước giải: Trả lời
Giải thích: `(1 – 1/(x+1))(1-1/(x+2))(1-1/(x+3))…(1-1/(x+2019))` `= x/(x+1) . (x+1)/(x+2) . (x+2)/(x+3) … (x + 2018)/(x+2019)` `= [x . (x + 1)(x + 2)…(x+2018)]/[(x+1)(x+2)…(x+2018) . (x+2019)]` `= x/(x + 2019)` `\text{Học tốt!!!}` Trả lời
Đáp án:
`(1 – 1/(x+1))(1-1/(x+2))(1-1/(x+3))…(1-1/(x+2019))`
`= x/(x+1) . (x+1)/(x+2) . (x+2)/(x+3) … (x + 2018)/(x+2019)`
`= [x . (x + 1)(x + 2)…(x+2018)]/[(x+1)(x+2)…(x+2018) . (x+2019)]`
`= x/(x + 2019)`
Giải thích các bước giải:
Giải thích:
`(1 – 1/(x+1))(1-1/(x+2))(1-1/(x+3))…(1-1/(x+2019))`
`= x/(x+1) . (x+1)/(x+2) . (x+2)/(x+3) … (x + 2018)/(x+2019)`
`= [x . (x + 1)(x + 2)…(x+2018)]/[(x+1)(x+2)…(x+2018) . (x+2019)]`
`= x/(x + 2019)`
`\text{Học tốt!!!}`